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Moment of Inertia of Cavendish pendulum

  1. Aug 19, 2007 #1

    I am trying to compute the moment of inertia of the Cavendish pendulum. I used

    I = 2(m r^2)

    r = gyration arm
    m = the weight attached to the pendulum

    But this formula is for a dumbell type of torsion pendulum where the weights are attached to the bar.

    Does anyone know the formula for a pendulum where the weights are suspended as in the case of the Cavendish pendulum? I couldn't find it online.

    With this formula I got

    I = 2 ( 729.8 * 93.1^2) = 12,651,243.56 g cm^2

    Does this sound right for a pendulum of this dimensions?

    Thanks for your help.
  2. jcsd
  3. Aug 19, 2007 #2


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    Staff: Mentor

    For the benefit of those of us who haven't been following your previous discussions closely, can you provide a diagram of the specific geometry that you have in mind, so we can see how it differs from Example 1 in the link you provided (which is the case where [itex]I = 2mr^2[/itex])?
  4. Aug 19, 2007 #3
    Thanks for your reply. Here's the drawing from Cavendish's original paper:


    And here's a detail


    Cavendish also gave the weight of the bar and the silver wire he used to strengthen it. I assume those will have to be taken into consideration as well.
  5. Aug 19, 2007 #4


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    Staff: Mentor

    You've still got two masses revolving around a common central point, so their moment of inertia I is the same as when they're connected by a rod of negligible mass in a dumbbell configuration as in the page you first referenced. So you're OK so far.

    What you need now is I for the rotating part of the suspension mechanism. You have the mass of the transverse rod, and I for a rod rotating transversely around its midpoint is a standard case. For some reason it's not on the page you first referenced, but you can find it on Hyperphysics:


    Just add the two I's together. In principle, you should also consider I for the rest of the rotating part of the apparatus, but the supporting wires are probably pretty light so you can safely neglect them, at least as a start in your analysis.
  6. Aug 19, 2007 #5
    Ok. Thanks. From hyperphysics I got 1/12 M L^2 and added it to the previous result:

    I = 2 m r^2 + 1/12 M L^2

    L = length of the rod
    r = half length of the rod


    Total moment of inertia for the Cavendish experiment = r^2 (2m + 1/3M) = 13,137,851.84 cm2 g

    m = weight of each ball = 729.8 g

    M = Rod weight = 155.5 + 11 + 2.9 = 169.40 g (including deal rod, silver supporting wire, vernier attached to the arm)

    L = Rod length = 73.3 inches = 186.18 cm

    r = Half rod length = 93.09 cm

    I also added another picture with a little more detail about the arm. (Figure 3 here.)

    Does this look good?

    Thanks again for your help.
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